# 327 Count of Range Sum

## 327. [Count of Range Sum](https://leetcode.com/problems/count-of-range-sum/description/)

## 1. Question

Given an integer array`nums`, return the number of range sums that lie in`[lower, upper]`inclusive.\
Range sum`S(i, j)`is defined as the sum of the elements in`nums`between indices`i`and`j`(`i`≤`j`), inclusive.

**Note:**\
A naive algorithm of O(n^2) is trivial. You MUST do better than that.

**Example:**\
Givennums=`[-2, 5, -1]`,lower=`-2`,upper=`2`,\
Return`3`.\
The three ranges are :`[0, 0]`,`[2, 2]`,`[0, 2]`and their respective sums are:`-2, -1, 2`.

## 2. Implementation

**(1) TreeMap**

我们定义sum\[i]为nums\[0...i]之间的subarray sum, 对于区间\[j,i]的subarray sum，其值等于sum\[i] - sum\[j], 如果 lower <= sum\[i] - sum\[j] <= upper, 通过推导可以得出 sum\[j] - upper <= sum\[j] <= sum\[i] - lower. 我们遍历数组的时候，利用sum变量加上当前的数，这样我们很容易得到sum\[i], 要得到sum\[j], 我们利用TreeMap的区间查询的特性，查询treemap，key在 sum\[i] - upper\[i] 和 sum\[i] - lower之间的值的个数，最后得到结果

```java
public class Solution {
    public int countRangeSum(int[] nums, int lower, int upper) {        
        TreeMap<Long, Long> map = new TreeMap();
        map.put((long)0, (long)1);
        long sum = 0;
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            long fromKey = sum - upper;
            long toKey = sum - lower;
            Map<Long, Long> subMap = map.subMap(fromKey, true, toKey, true);

            for (long value : subMap.values()) {
                count += value;
            }

            if (map.containsKey(sum)) {
                map.put(sum, map.get(sum) + 1);
            }
            else {
                map.put(sum, (long)1);
            }
        }
        return count;
    }
}
```

**(2) Merge Sort**

思路: 这里的思路和Leetcode 315很类似，我们利用merge sort，在merge的过程中count 符合条件的区间和个数. 具体在我们的merge()里，我们知道sum\[start, mid] 和 sum\[mid + 1, end]这两个区间是已经排好序的，对于每个在左区间的数sum\[left], 我们分别要找出一个index high满足sum\[high] - sum\[left] > upper, 以及第一个index low满足sum\[low] - sum\[left] >= lower, 这样符合条件的区间个数就是 high - low. 在计算符合条件的区间个数同时，我们也顺便完成merge的过程

```java
class Solution {
    public int countRangeSum(int[] nums, int lower, int upper) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        int n = nums.length;
        long[] sum = new long[n + 1];

        for (int i = 1; i <= n; i++) {
            sum[i] = sum[i - 1] + nums[i - 1];
        }

        int[] count = new int[1];
        mergeSort(sum, 0, sum.length - 1, lower, upper, count);
        return count[0];
    }

    public void mergeSort(long[] sum, int start, int end, int lower, int upper, int[] count) {
        if (start >= end) {
            return;
        }

        int mid = start + (end - start) / 2;
        mergeSort(sum, start, mid, lower, upper, count);
        mergeSort(sum, mid + 1, end, lower, upper, count);
        merge(sum, start, mid, end, lower, upper, count);
    }

    public void merge(long[] sum, int start, int mid, int end, int lower, int upper, int[] count) {
        long[] temp = new long[end - start + 1];

        int low = mid + 1;
        int high = mid + 1;
        int right = mid + 1;
        int tIndex = 0;

        for (int left = start; left <= mid; left++) {
            while (low <= end && sum[low] - sum[left] < lower) ++low;
            while (high <= end && sum[high] - sum[left] <= upper) ++high;

            count[0] += high - low;

            while (right <= end && sum[right] <= sum[left]) temp[tIndex++] = sum[right++];
            temp[tIndex++] = sum[left];
        }

        while (right <= end) {
            temp[tIndex++] = sum[right++];
        }

        for (int k = start; k <= end; k++) {
            sum[k] = temp[k - start];
        }
    }
}
```

**(3) Segment Tree**

```
```

## 3. Time & Space Complexity

**TreeMap**: 时间复杂度O(nlogn), 空间复杂度O(n)

**Merge Sort**: 时间复杂度O(nlogn), 空间复杂度O(n)


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://protegejj.gitbook.io/algorithm-practice/google/327-count-of-range-sum.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.